Here is the question I am trying to prove:
Let $M$ be a matroid, and $r$ and $cl$ be its rank function and its closure operator. Prove the following:
$(d)$ $r(X \cup Y) = r(X \cup cl(Y)) = r(cl(X) \cup cl(Y)) = r(cl(X \cup Y))$
Here is my attempt:
First, proving that $r(X \cup Y) = r(X \cup cl(Y)).$
Since $X \subseteq X$ and $Y \subseteq cl(Y),$ then $X \cup Y \subseteq X \cup cl(Y) $ and hence by the properties of the rank function we have $r(X \cup Y) \leq r(X \cup cl(Y)) $. Now, for proving the other direction I do not know how to get that $cl(Y) \subseteq Y.$ I know that $X \subseteq X$ and $r(Y) = r(cl(Y))$ but I do not know how to complete. Any help will be appreciated.
We have that the rank of a subset $Z$ of $M$ is the smallest cardinality of a subset $K$ of $M$ such that $Z\subset cl(K)$. It follows that for any subsets $Z$, $T$ of $M$ we have $r(Z)\le r(T)$ provided $Z\subset T$. Since the closure operator on $M$ is monotone and idempotent, if $Z\subset cl(K)$ then $cl(Z)\subset cl(cl(K))=cl(K)$, thus $r(Z)=r(cl(Z))$. The monotonicity of rank implies the following lemma.
Lemma. We have $r(Z)=r(T)=r(cl(Z))$ for any sets $T$, $Z\subset T\subset cl(T)$. $\square$
We have $X\subset cl(X)$, $Y\subset cl(Y)$. On the other hand, since $X\subset X\cup Y$ and $Y\subset X\cup Y$, we have $cl(X)\subset cl(X\cup Y)$ and $cl(Y)\subset cl(X\cup Y)$. Thus $$X\cup Y\subset X\cup cl(Y)\subset cl(X)\cup cl(Y)\subset cl(X\cup Y).$$ Lemma applied to $Z=X\cap Y$ and $T=X\cup cl(Y)$ or $T=cl(X)\cup cl(Y)$ implies that $$r(X\cup Y)=r(X\cup cl(Y))=r(cl(X)\cup cl(Y))=r(cl(X\cup Y)).$$